Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Second Edition Hello all I was presented with this question from Folland's real analysis second edition on Radon measures which I am stuck on and so would really appreciate the help on.
I m a novice in Radon measures especially in the concepts and abstractions so any help would be appreciated. It is problem #7 on page 220
It reads as follows:

Just in case, the definition of Radon Measure used in the book is:
A Radon measure on X is a Borel measure that is finite on all
compact sets, outer regular on all Borel sets, and inner regular on all open sets.
I also know for a fact that Radon measures are also inner regular on all their sigma finite sets.
I have tried to attack the problem many times but to no avail as for some reason I cannot seem to incorporate the given assumption that X is sigma finite.
Also the assumption is X is locally compact Hausdorff space.
I would really appreciate the help Thanks
EDIT: I figured it is obviously finite on compact sets
Possible direction I have:
I have studied in the Folland book that all $ \sigma-finite $ Radon measures are regular therefore for all Borel sets A and E we have $ \mu_A(E)=\mu(E \cap A) $ is the supremum on measure of all compact subsets in $ E \cap A $. How do I move further? I need inner regularity on open sets E meaning the supremum on measures of compact subsets of E not $ A \cap E $ as I have.
 A: Let's check whether $\mu_A$ satisfies the conditions of being a Radon measure step by step.
1. $\mu_A$ is a measure. This is easy, I leave it to you.
2. $\mu_A$ is finite on compact sets. Let $K\subseteq X$ be compact. Then, $$\mu_A(K)=\mu(K\cap A)\leq \mu(K)<\infty,$$ given that $\mu$ is a Radon measure.
3. $\mu_A$ is outer regular on all Borel sets. Let $E\in\mathscr B_X$. We want to show that $$\mu_A(E)=\inf\{\mu_A(U)\,|\,E\subseteq U\text{ and $U$ is open}\}.$$ The direction $\leq$ is easy to see, for if $E\subseteq U$ and $U$ is open, then $\mu_A(E)\leq\mu_A(U)$; now take infimum. The challenging part is the direction $\geq$. Fix $\varepsilon>0$. Since $\mu$ is $\sigma$-finite, there exists a sequence of $(F_n)_{n\in\mathbb N}$ of Borel-measurable sets such that $\mu(F_n)<\infty$ for all $n\in\mathbb N$ and $X=\bigcup_{n=1}^{\infty} F_n$. Without loss of generality, the $(F_n)_{n\in\mathbb N}$ can be taken to be disjoint. (Exercise: why?) Now, for any fixed $n\in\mathbb N$, one can find an open set $U_n\subseteq X$ such that 


*

*$E\cap F_n\subseteq U_n$; and

*$\mu(U_n)< \mu(E\cap F_n)+\varepsilon/2^n<\infty$,
given that $\mu$ is outer regular and $\mu(F_n)<\infty$. Note that
\begin{align*}
\mu(U_n\cap A)=&\,\mu(U_n\cap [E\cap F_n]^c\cap A)+\mu(U_n\cap E\cap F_n\cap A)\\
=&\,\mu(U_n\cap [E\cap F_n]^c\cap A)+\mu(E\cap F_n\cap A)\\
\leq&\,\mu(U_n\cap [E\cap F_n]^c)+\mu(E\cap F_n\cap A)\\
=&\,\mu(U_n)-\mu(E\cap F_n)+\mu(E\cap F_n\cap A)<\mu(E\cap F_n\cap A)+\varepsilon/2^n.
\end{align*}
Letting $U\equiv\bigcup_{n\in\mathbb N}U_n$, we have that $U$ is open, $$E=\bigcup_{n\in\mathbb N}E\cap F_n\subseteq\bigcup_{n\in\mathbb N} U_n=U,$$ and
\begin{align*}
\mu_A(U)=&\,\mu(U\cap A)\leq\sum_{n=1}^{\infty}\mu(U_n\cap A)\leq\sum_{n=1}^{\infty}\mu(E\cap F_n\cap A)+\sum_{n=1}^{\infty}\frac{\varepsilon}{2^n}\\
=&\,\mu\left(\bigcup_{n\in\mathbb N}E\cap F_n\cap A\right)+\varepsilon=\mu(E\cap A)+\varepsilon=\mu_A(E)+\varepsilon.
\end{align*}
Since $\varepsilon$ can be made arbitrarily small, it follows that $$\mu_A(E)\geq\inf\{\mu_A(U)\,|\,E\subseteq U\text{ and $U$ is open}\}.$$
4. $\mu_A$ is inner regular on open sets. The claim is that for any open subset $U\subseteq X$, $$\mu_A(U)=\sup\{\mu_A(K)\,|\,K\subseteq U\text{ and $K$ is compact}\}.$$ Again, the direction $\geq$ is clear. For the direction $\leq$, let's consider two cases.
Case I: $\mu(U\cap A)<\infty$. Fix $\varepsilon>0$. By the outer regularity of $\mu$, there exists some open set $V\subseteq X$ such that


*

*$U\cap A\subseteq V$; and

*$\mu(V)<\mu(U\cap A)+\varepsilon<\infty$.


Now, $U\cap V$ is an open set of finite measure. In turn, by the inner regularity of $\mu$, there exists some compact set $K\subseteq X$ such that


*

*$K\subseteq U\cap V$; and

*$\mu(K)>\mu(U\cap V)-\varepsilon$.


Therefore,
\begin{align*}
\mu_A(K)=&\,\mu(K\cap A)=\mu(U\cap V\cap A)-\mu(U\cap V\cap K^c\cap A)\\
=&\,\mu(U\cap A)-\mu(U\cap V\cap K^c)\\
=&\,\mu(U\cap A)-[\mu(U\cap V)-\mu(K)]\\
>&\,\mu(U\cap A)-\varepsilon=\mu_A(U)-\varepsilon,
\end{align*}
and also $K\subseteq U\cap V\subseteq U$. Therefore, $$\mu_A(U)\leq\sup\{\mu_A(K)\,|\,K\subseteq U\text{ and $K$ is compact}\}.$$
Case II: $\mu(U\cap A)=\infty$. Fix any number $C>0$. Let $(F_n)_{n\in\mathbb N}$ be as before. Since $$\infty=\mu(U\cap A)=\sum_{n=1}^{\infty}\mu(U\cap A\cap F_n),$$ there must exist some $N\in\mathbb N$ such that $$C+1<\sum_{n=1}^{N}\mu(U\cap A\cap F_n)=\mu\left[U\cap A\cap\left(\bigcup_{n=1}^N F_n\right)\right]<\infty.$$ Let $G\equiv\bigcup_{n=1}^N F_n$. Again, one can find an open set $V\subseteq X$ such that


*

*$U\cap A\cap G\subseteq V$; and

*$\mu(V)<\mu(U\cap A\cap G)+1<\infty$.


This way, $U\cap V$ is an open set of finite measure. In turn, we can find a compact set $K\subseteq X$ such that


*

*$K\subseteq U\cap V\subseteq U$; and

*$\mu(K)>\mu(U\cap V)-1$.


We now find that
\begin{align*}
\mu_A(K)=&\,\mu(K\cap A)=\mu(U\cap V\cap A)-\mu(U\cap V\cap K^c\cap A)\\
\geq&\,\mu(U\cap A\cap G)-\mu(U\cap V\cap K^c)\\
=&\,\mu(U\cap A\cap G)-[\mu(U\cap V)-\mu(K)]\\
>&\,(C+1)-1=C.
\end{align*}
In words, if $\mu_A(U)=\infty$, then $U$ contains compact sets of arbitrarily large $\mu_A$-measure. The proof is complete.
